Francois Viete

Francois Viete was born in 1540 in Fontenay le Comte, died on 23 February 1603. He was a French mathematician and astronomer.

He studied and practised law, then worked as a teacher of science in the homes of the nobility and royal adviser and parliamentary (the Parliament of Brittany).

He dealt with algebra and trigonometry, formulated algebra formulas, allowing to solve quadratic equations.

 

Vieta’s formulas

If the quadratic equation ax² + b + c = 0 has roots x1, x2, are:

  x1 + x2 = - b/a and x1 ∙ x2 = c/a

 x1, x2 – rootsa, b, c - coefficient equation

Was the first to consistently apply the letter symbols in algebra, although its notation is significantly different from currently used.

  

He was a secret adviser at the court of Henry III and Henry IV

During the French-Spanish war, he deserved to be breaking enemy codes.

If the equation has roots, calculate their sum and product.

2x2-20x + 15 = 0 Δ = 400-120 = 280 > 0 , so there are two roots x1 , x2 , x1 + x2 = - b/a = 20/2 = 10 , x1*x2 = c / a = 15/2

Task 1

Zadanie 1Jeśli równanie ma pierwiastki, oblicz

ich sumę i iloczyn.2x2-20x + 15 = 0 Δ = 400-120 = 280

> 0 , zatem istnieją dwa pierwiastki x1, x2 . x1 + x2 = - b / a = 20 /2 = 10 , x1* x2 = c / a = 15/2

Task 2If equation 6x2-9x + 2 = 0 has roots ,

calculate the sum of the squares ,Since Δ = 81-48 = 33 > 0 , the

equation has two roots x1 x2 . Note that :

x12+x2

2=(x1+x2)2-2x1x2=(-b/a)2-2*c/athen:x1

2+x22=(9/6)2-2*2/6=9/4-2/3=19/12

Zadanie 2 Jeśli równanie 6x2-9x+2=0 ma

pierwiastki, oblicz sumę ich kwadratów,

Ponieważ Δ=81-48=33>0, równanie ma dwa pierwiastki x1 x2. Zauważmy, że

x12+x2

2=(x1+x2)2-2x1x2=(-b/a)2-2*c/azatem:x1

2+x22=(9/6)2-2*2/6=9/4-2/3=19/12

Task 3

We use the abridged formula condensed the amount cubes. calculate

We Substitute Vieta's formulas and get

zadanie 3 Korzystamy ze wzoru skróconego

mnożenia na sumę sześcianów. Oblicz

Podstawiamy wzory Viete'a i otrzymujemy: